College Physics for AP® Courses

# 8.5Inelastic Collisions in One Dimension

College Physics for AP® Courses8.5 Inelastic Collisions in One Dimension

### Learning Objectives

By the end of this section, you will be able to:

• Define inelastic collision.
• Explain perfectly inelastic collisions.
• Apply an understanding of collisions to sports.
• Determine recoil velocity and loss in kinetic energy given mass and initial velocity.

The information presented in this section supports the following AP® learning objectives and science practices:

• 4.B.1.1 The student is able to calculate the change in linear momentum of a two-object system with constant mass in linear motion from a representation of the system (data, graphs, etc.). (S.P. 1.4, 2.2)
• 5.A.2.1 The student is able to define open and closed systems for everyday situations and apply conservation concepts for energy, charge, and linear momentum to those situations. (S.P. 6.4, 7.2)
• 5.D.1.3 The student is able to apply mathematical routines appropriately to problems involving elastic collisions in one dimension and justify the selection of those mathematical routines based on conservation of momentum and restoration of kinetic energy. (S.P. 2.1, 2.2)
• 5.D.1.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2)
• 5.D.2.1 The student is able to qualitatively predict, in terms of linear momentum and kinetic energy, how the outcome of a collision between two objects changes depending on whether the collision is elastic or inelastic. (S.P. 6.4, 7.2)
• 5.D.2.2 The student is able to plan data collection strategies to test the law of conservation of momentum in a two-object collision that is elastic or inelastic and analyze the resulting data graphically. (S.P.4.1, 4.2, 5.1)
• 5.D.2.3 The student is able to apply the conservation of linear momentum to a closed system of objects involved in an inelastic collision to predict the change in kinetic energy. (S.P. 6.4, 7.2)
• 5.D.2.4 The student is able to analyze data that verify conservation of momentum in collisions with and without an external friction force. (S.P. 4.1, 4.2, 4.4, 5.1, 5.3)
• 5.D.2.5 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum as the appropriate solution method for an inelastic collision, recognize that there is a common final velocity for the colliding objects in the totally inelastic case, solve for missing variables, and calculate their values. (S.P. 2.1 2.2)
• 5.D.2.6 The student is able to apply the conservation of linear momentum to an isolated system of objects involved in an inelastic collision to predict the change in kinetic energy. (S.P. 6.4, 7.2)

We have seen that in an elastic collision, internal kinetic energy is conserved. An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). This lack of conservation means that the forces between colliding objects may remove or add internal kinetic energy. Work done by internal forces may change the forms of energy within a system. For inelastic collisions, such as when colliding objects stick together, this internal work may transform some internal kinetic energy into heat transfer. Or it may convert stored energy into internal kinetic energy, such as when exploding bolts separate a satellite from its launch vehicle.

### Inelastic Collision

An inelastic collision is one in which the internal kinetic energy changes (it is not conserved).

Figure 8.11 shows an example of an inelastic collision. Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially $1 2 m v 2 + 1 2 m v 2 =m v 2 1 2 m v 2 + 1 2 m v 2 =m v 2$. The two objects come to rest after sticking together, conserving momentum. But the internal kinetic energy is zero after the collision. A collision in which the objects stick together is sometimes called a perfectly inelastic collision because it reduces internal kinetic energy more than does any other type of inelastic collision. In fact, such a collision reduces internal kinetic energy to the minimum it can have while still conserving momentum.

### Perfectly Inelastic Collision

A collision in which the objects stick together is sometimes called “perfectly inelastic.”

Figure 8.11 An inelastic one-dimensional two-object collision. Momentum is conserved, but internal kinetic energy is not conserved. (a) Two objects of equal mass initially head directly toward one another at the same speed. (b) The objects stick together (a perfectly inelastic collision), and so their final velocity is zero. The internal kinetic energy of the system changes in any inelastic collision and is reduced to zero in this example.

### Example 8.5

#### Calculating Velocity and Change in Kinetic Energy: Inelastic Collision of a Puck and a Goalie

(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See Figure 8.12 )

Figure 8.12 An ice hockey goalie catches a hockey puck and recoils backward. The initial kinetic energy of the puck is almost entirely converted to thermal energy and sound in this inelastic collision.

#### Strategy

Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested.

#### Solution for (a)

Momentum is conserved because the net external force on the puck-goalie system is zero.

Conservation of momentum is

$p 1 + p 2 = p′ 1 + p′ 2 p 1 + p 2 = p′ 1 + p′ 2 size 12{p rSub { size 8{1} } +p rSub { size 8{2} } = { {p}} sup { ' } rSub { size 8{1} } + { {p}} sup { ' } rSub { size 8{2} } } {}$
8.63

or

$m 1 v 1 + m 2 v 2 = m 1 v′ 1 + m 2 v′ 2 . m 1 v 1 + m 2 v 2 = m 1 v′ 1 + m 2 v′ 2 . size 12{m rSub { size 8{1} } v rSub { size 8{1} } +m rSub { size 8{2} } v rSub { size 8{2} } =m rSub { size 8{1} } { {v}} sup { ' } rSub { size 8{1} } +m rSub { size 8{2} } { {v}} sup { ' } rSub { size 8{2} } } {}$
8.64

Because the goalie is initially at rest, we know $v2=0v2=0 size 12{v rSub { size 8{2} } =0} {}$. Because the goalie catches the puck, the final velocities are equal, or $v′ 1 = v′ 2 =v′ v′ 1 = v′ 2 =v′ size 12{ { {v}} sup { ' } rSub { size 8{1} } = { {v}} sup { ' } rSub { size 8{2} } =v'} {}$. Thus, the conservation of momentum equation simplifies to

$m1v1=m1+m2v′.m1v1=m1+m2v′. size 12{m rSub { size 8{1} } v rSub { size 8{1} } = left (m rSub { size 8{1} } +m rSub { size 8{2} } right )v'} {}$
8.65

Solving for $v′v′ size 12{v'} {}$ yields

$v′=m1m1+m2v1.v′=m1m1+m2v1. size 12{v'= { {m rSub { size 8{1} } } over {m rSub { size 8{1} } +m rSub { size 8{2} } } } v rSub { size 8{1} } } {}$
8.66

Entering known values in this equation, we get

$v′ = 0.150 kg 70.0 kg + 0.150 kg 35 .0 m/s = 7 . 48 × 10 − 2 m/s . v′ = 0.150 kg 70.0 kg + 0.150 kg 35 .0 m/s = 7 . 48 × 10 − 2 m/s . size 12{v'= left ( { {0 "." "150""kg"} over {"70" "." 0"kg"+0 "." "150""kg"} } right ) left ("35" "." 0"m/s" right )=7 "." "48" times "10" rSup { size 8{ - 2} } "m/s" "." } {}$
8.67

#### Discussion for (a)

This recoil velocity is small and in the same direction as the puck’s original velocity, as we might expect.

#### Solution for (b)

Before the collision, the internal kinetic energy $KEintKEint size 12{"KE" rSub { size 8{"int"} } } {}$ of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, $KEintKEint size 12{"KE" rSub { size 8{"int"} } } {}$ is initially

$KE int = 1 2 mv 2 = 1 2 0 . 150 kg 35 .0 m/s 2 = 91 . 9 J . KE int = 1 2 mv 2 = 1 2 0 . 150 kg 35 .0 m/s 2 = 91 . 9 J .$
8.68

After the collision, the internal kinetic energy is

$KE′ int = 1 2 m + M v 2 = 1 2 70 . 15 kg 7 . 48 × 10 − 2 m/s 2 = 0.196 J. KE′ int = 1 2 m + M v 2 = 1 2 70 . 15 kg 7 . 48 × 10 − 2 m/s 2 = 0.196 J.$
8.69

The change in internal kinetic energy is thus

$KE′ int − KE int = 0.196 J − 91.9 J = − 91.7 J KE′ int − KE int = 0.196 J − 91.9 J = − 91.7 J$
8.70

#### Discussion for (b)

Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. $KEintKEint size 12{"KE" rSub { size 8{"int"} } } {}$ is mostly converted to thermal energy and sound.

During some collisions, the objects do not stick together and less of the internal kinetic energy is removed—such as happens in most automobile accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. Figure 8.13 shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example 8.6 deals with data from such a collision.

Figure 8.13 An air track is nearly frictionless, so that momentum is conserved. Motion is one-dimensional. In this collision, examined in Example 8.6, the potential energy of a compressed spring is released during the collision and is converted to internal kinetic energy.

Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports—a lightweight bat (such as a softball bat) cannot hit a hardball very far.

The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the “sweet spot” on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.

### Take-Home Experiment—Bouncing of Tennis Ball

1. Find a racquet (a tennis, badminton, or other racquet will do). Place the racquet on the floor and stand on the handle. Drop a tennis ball on the strings from a measured height. Measure how high the ball bounces. Now ask a friend to hold the racquet firmly by the handle and drop a tennis ball from the same measured height above the racquet. Measure how high the ball bounces and observe what happens to your friend’s hand during the collision. Explain your observations and measurements.
2. The coefficient of restitution $(c)(c) size 12{ $$c$$ } {}$ is a measure of the elasticity of a collision between a ball and an object, and is defined as the ratio of the speeds after and before the collision. A perfectly elastic collision has a $cc size 12{c} {}$ of 1. For a ball bouncing off the floor (or a racquet on the floor), $cc size 12{c} {}$ can be shown to be $c=(h/H)1/2c=(h/H)1/2 size 12{c= $$h/H$$ rSup { size 8{1/2} } } {}$ where $hh size 12{h} {}$ is the height to which the ball bounces and $HH size 12{H} {}$ is the height from which the ball is dropped. Determine $cc size 12{c} {}$ for the cases in Part 1 and for the case of a tennis ball bouncing off a concrete or wooden floor ($c=0.85c=0.85 size 12{c=0 "." "85"} {}$ for new tennis balls used on a tennis court).

### Example 8.6

#### Calculating Final Velocity and Energy Release: Two Carts Collide

In the collision pictured in Figure 8.13, two carts collide inelastically. Cart 1 (denoted $m1m1 size 12{m rSub { size 8{1} } } {}$ carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of $2.00 m/s2.00 m/s size 12{2 "." "00""m/s"} {}$. Cart 2 (denoted $m2m2 size 12{m rSub { size 8{2} } } {}$ in Figure 8.13) has a mass of 0.500 kg and an initial velocity of $−0.500 m/s−0.500 m/s size 12{ - 0 "." "500""m/s"} {}$. After the collision, cart 1 is observed to recoil with a velocity of $−4.00 m/s−4.00 m/s size 12{ - 4 "." "00""m/s"} {}$. (a) What is the final velocity of cart 2? (b) How much energy was released by the spring (assuming all of it was converted into internal kinetic energy)?

#### Strategy

We can use conservation of momentum to find the final velocity of cart 2, because $Fnet=0Fnet=0 size 12{F rSub { size 8{"net"} } =0} {}$ (the track is frictionless and the force of the spring is internal). Once this velocity is determined, we can compare the internal kinetic energy before and after the collision to see how much energy was released by the spring.

#### Solution for (a)

As before, the equation for conservation of momentum in a two-object system is

$m 1 v 1 + m 2 v 2 = m 1 v′ 1 + m 2 v′ 2 . m 1 v 1 + m 2 v 2 = m 1 v′ 1 + m 2 v′ 2 .$
8.71

The only unknown in this equation is $v′ 2 v′ 2$. Solving for $v′ 2 v′ 2$ and substituting known values into the previous equation yields

$v′ 2 = m 1 v 1 + m 2 v 2 − m 1 v′ 1 m 2 = 0.350 kg 2.00 m/s + 0.500 kg − 0.500 m/s 0.500 kg − 0.350 kg − 4.00 m/s 0.500 kg = 3.70 m/s. v′ 2 = m 1 v 1 + m 2 v 2 − m 1 v′ 1 m 2 = 0.350 kg 2.00 m/s + 0.500 kg − 0.500 m/s 0.500 kg − 0.350 kg − 4.00 m/s 0.500 kg = 3.70 m/s.$
8.72

#### Solution for (b)

The internal kinetic energy before the collision is

$KEint = 12m1 v12+12 m2 v22 = 120.350 kg2.00 m/s2+120.500 kg–0.500 m/s2 = 0.763 J. KEint = 12m1 v12+12 m2 v22 = 120.350 kg2.00 m/s2+120.500 kg–0.500 m/s2 = 0.763 J.$
8.73

After the collision, the internal kinetic energy is

$KE′int = 12m1v′12+12 m2v′22 = 1 2 0.350 kg - 4.00 m/s 2 + 1 2 0.500 kg 3.70 m/s 2 = 6.22 J. KE′int = 12m1v′12+12 m2v′22 = 1 2 0.350 kg - 4.00 m/s 2 + 1 2 0.500 kg 3.70 m/s 2 = 6.22 J.$
8.74

The change in internal kinetic energy is thus

$KE′ int − KE int = 6.22 J − 0 . 763 J = 5.46 J. KE′ int − KE int = 6.22 J − 0 . 763 J = 5.46 J.$
8.75

#### Discussion

The final velocity of cart 2 is large and positive, meaning that it is moving to the right after the collision. The internal kinetic energy in this collision increases by 5.46 J. That energy was released by the spring.

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